Question

Of all the closed cylindrical cans (right circular), of a given volume of $100$ cubic centimetres, find the dimensions of the can which has the minimum surface area?

Answer 1

Volume $=$ $\pi r^{2}h=100$

Total surface area is given by $2\pi rh+2\pi r^2$

$=$ $\frac{200}{r}+2\pi r^2$

Now let's maximise this

$\frac{ds}{dr}=\frac{-200+4\pi r^3}{r^2}$

Putting this to zero we get

$4\pi r^3-200=0$

$r={\left(\frac{50}{\pi }\right)}^{\frac{1}{3}}$

Now lets look at the second derivative

$\frac{d^{2}s}{d{r}^2}=\frac{400}{r^3}+4\pi$

And this is positive at critical $r$ so we will have minimum area at this $r$

So height is $\frac{100}{\pi }{\left(\frac{\pi }{50}\right)}^{\frac{2}{3}}$